Q12 / 155 · WBJEE 2025 · Chemistry
${ }_5 \mathrm{~B}^{10}+{ }_2 \mathrm{He}^4 \rightarrow \mathrm{X}+{ }_{\mathrm{o}} \mathrm{n}^1$
In the above nuclear reaction ' X ' will be
- A${ }_7 \mathrm{~N}^{14}$
- Correct answerB${ }_7 \mathrm{~N}^{13}$
- C${ }_6 \mathrm{C}^{12}$
- D${ }_7 \mathrm{~N}^{14}$
Solution
${ }_5 \mathrm{~B}^{10}+{ }_2 \mathrm{He}^4 \longrightarrow{ }_b^a \mathrm{x}+{ }_0 \mathrm{n}^1$
Balancing atomic numbers, we have
$\begin{aligned}
& 5+2=\mathrm{b}+0 \\
& \therefore \mathrm{~b}=7
\end{aligned}$
Balancing mass numbers; we have
$\begin{aligned}
& 10+4=a+1 \\
& \therefore a=13
\end{aligned}$
Hence, the correct answer is option: 'B'.